Existence and uniqueness for projected dynamics and convergence of projection Euler under countable and dependent choice

Since $d\ge 1$, [F5] shows that compact $C$ is closed and bounded. Along with the stated nonemptiness and convexity, this verifies the set hypotheses in [F1] and the projection suppliers before they are used. Boundedness gives a finite diameter bound $D$ for $C$. Put $M=\mathrm{\parallel }f(0,x_0){\mathrm{\parallel }}_2+L(T+D)$; the Lipschitz hypothesis gives $\mathrm{\parallel }f(t,x){\mathrm{\parallel }}_2\le M$ on $[0,T]\times C$. For each exact Euler step, $P_C(x_k)=x_k$, so [F2] gives $\mathrm{\parallel }{x}_{k+1}-x_k{\mathrm{\parallel }}_2\le h\mathrm{\parallel }f(t_k,x_k){\mathrm{\parallel }}_2\le hM$. Convexity keeps the affine interpolant in $C$, and its slope on each step has norm at most $M$.

On each open mesh cell $[t_k,t_{k+1})$, write $X_h$ for the affine interpolant, ${\bar{X}}_h=x_{k+1}$, $F_h=f(t_k,x_k)$, and $n_h=F_h-{\dot{X}}_h=(x_k+h{F}_h-x_{k+1})\mathrm{/}h$. By the projection variational inequality in [F1], $n_h\in N_C({\bar{X}}_h)$. Consequently, for every fixed $z\in C$, $⟨{\dot{X}}_h-F_h,z-{\bar{X}}_h⟩\ge 0.$

Apply [F3] to the sequence of interpolants for $N=1,2,\dots$. Along a subsequence $N_j\to \mathrm{\infty }$, they converge uniformly to a map $x:[0,T]\to C$ with $x(0)=x_0$ and Lipschitz constant $M$. By [F6], $x$ is absolutely continuous and differentiable almost everywhere; at each differentiability point, the Lipschitz inequality implies $\mathrm{\parallel }\dot{x}(t){\mathrm{\parallel }}_2\le M$.

Fix $0\le a<b\le T$ and a fixed $z\in C$. We have $\mathrm{\parallel }{\bar{X}}_h-X_h{\mathrm{\parallel }}_2\le Mh$ and $\mathrm{\parallel }{\dot{X}}_h-F_h{\mathrm{\parallel }}_2\le 2M$, so replacing ${\bar{X}}_h$ by $X_h$ in the integral of step 1.2 changes it by at most $2M^{2}h(b-a)$. The term with ${\dot{X}}_h$ is the endpoint difference of ${\mathrm{\Phi }}_z(X_h)$, where ${\mathrm{\Phi }}_z(w)=⟨w,z⟩-\frac{1}{2}\mathrm{\parallel }w{\mathrm{\parallel }}_2^2$. Uniform convergence gives its limit as the endpoint difference of ${\mathrm{\Phi }}_z(x)$; [F7] identifies that difference with ${\int }_a^b⟨\dot{x},z-x⟩$. Also $\mathrm{\parallel }{F}_h(t)-f(t,x(t)){\mathrm{\parallel }}_2\le L(h+Mh+\mathrm{\parallel }{X}_h-x{\mathrm{\parallel }}_{\mathrm{\infty }})$ on each mesh cell, so $F_h\to f(t,x(t))$ uniformly along the selected subsequence. Passing to the limit in the integrated inequality yields ${\int }_a^b{g}_z(t)dt\ge 0,g_z(t):=⟨\dot{x}(t)-f(t,x(t)),z-x(t)⟩\mathrm{.}$

The function $g_z$ is Lebesgue-integrable: [F7] gives a measurable $L^1$ version of $\dot{x}$; define it to be $0$ where the derivative is undefined. Boundedness of $f,x,z$ makes the other factors integrable. Extend this representative of $g_z$ by zero outside $[0,T]$ to obtain a locally integrable function on $\mathbb{R}$. Since every interval integral in step 3.1 is nonnegative, $G_z(t)={\int }_0^t{g}_z(s)ds$ is nondecreasing. At almost every interior $t$, [F8] gives ${\lim }_{r\downarrow 0}\frac{1}{2r}{\int }_{t-r}^{t+r}{g}_z(s)ds=g_z(t)\mathrm{.}$ For every sufficiently small $r>0$, the average on the left is $(G_z(t+r)-G_z(t-r))\mathrm{/}(2r)\ge 0$ because $G_z$ is nondecreasing. Taking the limit yields $g_z(t)\ge 0$ almost everywhere for each fixed $z\in C$.

Fix a bijection $\rho :\mathbb{N}\to \mathbb{Q}$ from [F9]. Define ${\pi }_1(n_1)=n_1$ and, recursively, ${\pi }_{k+1}(n_1,\dots ,n_{k+1})=J({\pi }_k(n_1,\dots ,n_k),n_{k+1})$. Since $J$ is a bijection, induction using [F10] shows that each ${\pi }_k:{\mathbb{N}}^k\to \mathbb{N}$ is a bijection. In particular, decoding $n$ by ${\pi }_d^{-1}$ and applying $\rho$ coordinatewise gives a surjection $s:\mathbb{N}\to {\mathbb{Q}}^d$. Thus $D_0=\{P_C(s(n)):n\in \mathbb{N}\}$ is countable. It is dense in $C$: for $z\in C$ and $\eta >0$, choose in each of the finitely many coordinates a rational $q_j$ strictly between $z_j-\eta \mathrm{/}(2d)$ and $z_j+\eta \mathrm{/}(2d)$, using the interval-density conclusion of [F10]. For $q=(q_1,\dots ,q_d)$, [F10] gives $\mathrm{\parallel }q-z{\mathrm{\parallel }}_2^2={\sum }_{j<d}\mathrm{\mid }{q}_j-z_j{\mathrm{\mid }}^2<d{\left(\frac{\eta }{2d}\right)}^2\le \frac{{\eta }^2}{4},$ so $\mathrm{\parallel }q-z{\mathrm{\parallel }}_2<\eta \mathrm{/}2$. Since $P_C(z)=z$, nonexpansiveness [F2] gives $\mathrm{\parallel }{P}_C(q)-z{\mathrm{\parallel }}_2=\mathrm{\parallel }{P}_C(q)-P_C(z){\mathrm{\parallel }}_2\le \mathrm{\parallel }q-z{\mathrm{\parallel }}_2<\eta$. For each $n$, put $z_n=P_C(s(n))\in D_0$. Step 4.1 gives a null exceptional set where $g_{z_n}<0$. Countable Choice [A1] and [F11] make their union, together with the null set where $\dot{x}$ does not exist, null. Outside it, $g_z(t)\ge 0$ for every $z\in D_0$. For each such $t$, the expression $g_z(t)$ is continuous in $z$; if it were negative at some $z\in C$, it would be negative on a neighborhood of $z$ meeting the dense set $D_0$, a contradiction. Hence the inequality holds for every $z\in C$, so $f(t,x(t))-\dot{x}(t)\in N_C(x(t))$ almost everywhere by [F1]. This proves existence of a solution along the selected subsequence.

The uniqueness lemma [F4], under the same choice assumptions through its FTC supplier, shows that every uniformly convergent subsequence limit of the Euler interpolants is this same solution $x$. If the full sequence did not converge uniformly to $x$, some subsequence would stay at uniform distance at least $\delta >0$; [F3] would give a further uniformly convergent subsequence, whose limit is a solution by steps 1.2–5.1 and hence equals $x$, a contradiction. Therefore the entire sequence converges uniformly to $x$.